At constant temperature 80% AB dissociates into  $A_2$ and $B_2$ , then the equilibrium constant for $2A{B}_{\left(g\right)}\rightleftharpoons A_{2\left(g\right)}+B_{2\left(g\right)}$  is

$\begin{array}{r}2A{B}_{\left(g\right)}\rightleftharpoons A_{2\left(g\right)}+B_{2\left(g\right)}\\ \text{ Let }{\mathrm{n}}_{\mathrm{AB}}=100\end{array}$

Percentage of decompositions of AB = 80

2 moles of AB decomposes 1 mole of ${\mathrm{A}}_2$ and 40 moles of $B_2$

$\begin{array}{r}\begin{array}{llll}\text{ Equilibrium concentration }& \frac{20}{V}& & \frac{40}{V}\\ K_C=\frac{\left[A_2\right]\left[B_2\right]}{[AB{]}^2}& \frac{40}{V}& & \end{array}\\ K_C=\frac{\left(\frac{40}{V}\right)\left(\frac{40}{V}\right)}{{\left(\frac{20}{V}\right)}^2}\\ K_C=\frac{40\times 40}{20\times 20}\\ K_C=4\\ \end{array}$